2003 IMO Shortlist S22

Let $n$ be a positive integer and let $(x_1,\ldots,x_n)$ , $(y_1,\ldots,y_n)$ be two sequences of positive real numbers. Suppose $(z_2,\ldots,z_{2n})$ is a sequence of positive real numbers such that $z_{i+j}^2 \geq x_iy_j$ for all $1\le i,j \leq n$ .

Let $M=\max\{z_2,\ldots,z_{2n}\}$ . Prove that $$\left( \frac{M+z_2+\dots+z_{2n}}{2n} \right)^2$$
\ge
\left( \frac{x_1+\dots+x_n}{n} \right)
\left( \frac{y_1+\dots+y_n}{n} \right).

<details><summary>comment</summary>*Edited by Orl.*</details>

*Proposed by Reid Barton, USA*

令 $n$ 为正整数，并令 $(x_1,\ldots,x_n)$ 、 $(y_1,\ldots,y_n)$ 为两个正实数序列。假设 $(z_2,\ldots,z_{2n})$ 是正实数序列，使得 $z_{i+j}^2 \geq x_iy_j$ 对于所有 $1\le i,j \leq n$ 。

设 $M=\max\{z_2,\ldots,z_{2n}\}$ 。证明$$

\left(\frac{M+z_2+\dots+z_{2n}}{2n} \right)^2

\ge

\left(\frac{x_1+\dots+x_n}{n} \right)

\left(\frac{y_1+\dots+y_n}{n}\right)。 $$

<details><summary>评论</summary>*由 Orl 编辑。*</details>

*由美国 Reid Barton 提出*